Question

the quadratic polynomial whose zeros are 1/2 and 1/3

Answer 1

$\alpha = \frac{1}{2 } \: and \: \: \beta = \frac{1}{3} \\ Sum \: of \: zeroes = \: (\alpha + \beta) \: = \: \frac{1}{2} + \: \frac{1}{3} = \frac{5}{6} \\ Product \: of \: zeroes = \: \alpha \beta = \: \frac{1}{2} × \: \frac{1}{3} = \frac{1}{6} \\ For \: quadratic \: polynomial\: \\ x^2 -(Sum\: of \: zeroes)x +(Product \: of \: zeroes) = 0 \\ x ^{2} - \frac{5x}{6} + \frac{1}{6} = 0 \\ 6x ^{2} - 5x + 1 = 0 \\so \: the \: quadratic \: polynomal\: is \: 6x^{2} - 5x + 1$

Answer 2

Given:- the quadratic polynomial whose zeros are 1/2 and 1/3

To find out:- the quadratic polynomial

According to the question:-

$\alpha = \frac{1}{2} \\\beta = \frac{1}{3}$

Now,

The formula of quadratic polynomial = $x^{2} + (sum of zeroes)x + (product of zeroes)=0\\x^{2} +(\alpha +\beta )x + (\alpha *\beta )=0$

Sum of Zeroes

$=\alpha + \beta \\= \frac{1}{2} + \frac{1}{3} \\=\frac{5}{6}$

Product of Zeroes

$=\alpha *\beta \\=\frac{1}{2} * \frac{1}{3} \\=\frac{1}{6}$

Put, the sum of zeroes and the product of zeroes in the formula

$x^{2} + (\alpha +\beta )x+ (\alpha *\beta )= 0\\x^{2} +\frac{5}{6} x + \frac{1}{6} = 0$

The quadratic polynomial whose zeros are $\frac{1}{2}$ and $\frac{1}{3}$ is $x^{2} +\frac{5}{6} x + \frac{1}{6} = 0$.

Learn More:-

Quadratic Polynomial:-

1. https://brainly.in/question/4030571

2. https://brainly.in/question/1150339

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